Simplify and expand the following expression: $ \dfrac{4}{3k + 18}- \dfrac{1}{3k + 27}- \dfrac{k}{k^2 + 15k + 54} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3k + 18} = \dfrac{4}{3(k + 6)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{1}{3k + 27} = \dfrac{1}{3(k + 9)}$ We can factor the quadratic in the third term: $ \dfrac{k}{k^2 + 15k + 54} = \dfrac{k}{(k + 6)(k + 9)}$ Now we have: $ \dfrac{4}{3(k + 6)}- \dfrac{1}{3(k + 9)}- \dfrac{k}{(k + 6)(k + 9)} $ The least common multiple of the denominators is: $ 9(k + 6)(k + 9)$ In order to get the first term over $9(k + 6)(k + 9)$ , multiply by $\dfrac{3(k + 9)}{3(k + 9)}$ $ \dfrac{4}{3(k + 6)} \times \dfrac{3(k + 9)}{3(k + 9)} = \dfrac{12(k + 9)}{9(k + 6)(k + 9)} $ In order to get the second term over $9(k + 6)(k + 9)$ , multiply by $\dfrac{3(k + 6)}{3(k + 6)}$ $ \dfrac{1}{3(k + 9)} \times \dfrac{3(k + 6)}{3(k + 6)} = \dfrac{3(k + 6)}{9(k + 6)(k + 9)} $ In order to get the third term over $9(k + 6)(k + 9)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{k}{(k + 6)(k + 9)} \times \dfrac{9}{9} = \dfrac{9k}{9(k + 6)(k + 9)} $ Now we have: $ \dfrac{12(k + 9)}{9(k + 6)(k + 9)} - \dfrac{3(k + 6)}{9(k + 6)(k + 9)} - \dfrac{9k}{9(k + 6)(k + 9)} $ $ = \dfrac{ 12(k + 9) - 3(k + 6) - 9k} {9(k + 6)(k + 9)} $ Expand: $ = \dfrac{12k + 108 - 3k - 18 - 9k}{9k^2 + 135k + 486} $ $ = \dfrac{90}{9k^2 + 135k + 486}$ Simplify: $ = \dfrac{10}{k^2 + 15k + 54}$